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Math Help - One to One

  1. #1
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    One to One

    If two sets X and Y have the same cardinality, can we assume that the f: X->Y is one to one? If not, and all Im given is that they're the same size, how can I prove that the function is injective? Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: One to One

    Quote Originally Posted by jzellt View Post
    If two sets X and Y have the same cardinality, can we assume that the f: X->Y is one to one? If not, and all Im given is that they're the same size, how can I prove that the function is injective? Thanks.
    If you know that |X|=|Y| then you can FIND an injection X\to Y. Give us more context.
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  3. #3
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    Re: One to One

    Other than the fact that the sets are finite with distinct elements, thats all Im given.
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  4. #4
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    Re: One to One

    I actually have to show that the function is bijective. I assumed that I must show injective and surjective. If there's a different approach, I'm all ears... Thanks
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  5. #5
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    Re: One to One

    Are you given any information about f? Since |X| = |Y|, there /exists/ a bijective function between these sets. But there also exist functions which are neither surjective nor injective (assuming that they have more than one element).

    Or maybe the assignment is to find such an f?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Re: One to One

    Quote Originally Posted by jzellt View Post
    I actually have to show that the function is bijective. I assumed that I must show injective and surjective. If there's a different approach, I'm all ears... Thanks
    Well, that's pretty important haha. It's true that a mapping f:X\to Y where |X|=|Y|<\infty is injective if and only if it's surjective. Does that help? (a proof can be found on my blog)
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